Question

A sample of natural rubber (200.0 g) is vulcanized, with the complete consumption of 4.8 g of sulfur. Natural rubber is a polymer of isoprene (C5H8). Four sulfur atoms are used in each crosslink connection. What percent of the isoprene units will be crosslinked

Answer 1

Answer: 1.3% many crosslinks as isoprene units,

Step-by-step explanation:

Given:

mass pf natural rubber= 200.0g

mass of sulphur = 4.8g

molar mass of sulphur =32g/mol

molar mass of isoprene = C5H8=( 12x5) +(1x8)= 68g/mol

Solution: we first find no of moles present in each using

no of moles =
`(mass)/(molarmass)`

Isoprene: 200.0g x [1mole / 68g] = 2.94moles.

Sulfur: 4.8g x [1mole / 32g] x [1 mole crosslinks / 4 moles S] = 0.0375 moles crosslinks.

to find % crosslinked units, we have

0.0375 / 2.94 = 1.3% as many crosslinks as isoprene units,